Answer:
A customer who sends 78 messages per day would be at 99.38th percentile.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Average of 48 texts per day with a standard deviation of 12.
This means that 
a. A customer who sends 78 messages per day would correspond to what percentile?
The percentile is the p-value of Z when X = 78. So



has a p-value of 0.9938.
0.9938*100% = 99.38%.
A customer who sends 78 messages per day would be at 99.38th percentile.
Answer:

Step-by-step explanation:
We are going to apply algebra to solve for the exponential growth
we know that the expression for the exponential formula is

where
a = the amount before measuring growth or decay
= 10,000
r = growth or decay rate = 23%= 0.23
x = number of time intervals that have passed
since this problem is on the growth we are going to replace "b" with (1+r), likewise for decay (1-r)
hence the formula becomes


The expression that will give the population value is

Answer:
Step-by-step explanation:
Answer:
Option A
Step-by-step explanation:
I am assuming that '3' is an exponent.
![x^3=64\\\\\sqrt[3]{x}=\sqrt[3]{64}\\\\\boxed{x=4}](https://tex.z-dn.net/?f=x%5E3%3D64%5C%5C%5C%5C%5Csqrt%5B3%5D%7Bx%7D%3D%5Csqrt%5B3%5D%7B64%7D%5C%5C%5C%5C%5Cboxed%7Bx%3D4%7D)
Hope this helps.
Answer:
I think that's right not 100% but hope it helps? xx